Strange error: cannot convert from 'int' to 'ios_base::openmode'

Question

I am using g++ to compile some code. I wrote the following snippet:

bool WriteAccess = true;
string Name = "my_file.txt";
ofstream File;
ios_base::open_mode Mode = std::ios_base::in | std::ios_base::binary;
if(WriteAccess)
  Mode |= std::ios_base::out | std::ios_base::trunc;
File.open(Name.data(), Mode);

And I receive these errors... any idea why?

Error 1: invalid conversion from ‘int’ to ‘std::_Ios_Openmode’
Error 2: initializing argument 2 of ‘std::basic_filebuf<_CharT, _Traits>* std::basic_filebuf<_CharT, _Traits>::open(const char*, std::_Ios_Openmode) [with _CharT = char, _Traits = std::char_traits]’

As far as I could tell from a Google search, g++ is actually breaking the C++ standard here. Which I find quite astonishing, since they generally conform very strictly to the standard. Is this the case? Or am I doing something wrong.

My reference for the standard: http://www.cplusplus.com/reference/iostream/ofstream/open/

Answer 1

openmode is the correct type, not open_mode.

Answer 2

This:

ios_base::open_mode Mode = std::ios_base::in | std::ios_base::binary;

should be:

std::ios_base::openmode Mode = std::ios_base::in | std::ios_base::binary;

Note the lack of _ in openmode.

(I had to add these lines and put your code in a function to get your snippet to compile.

#include <string>
#include <fstream>

using std::string;
using std::ofstream;
using std::ios_base;

)

Answer 3

g++ is not totally conforming, but it's not the reason for the error here.

The type of mode should be

std::ios_base::openmode

instead of

std::ios_base::open_mode

The latter is an old, deprecated API. It is still specified in Annex D of the C++ standard, so the compiler still have to support it.